Formulas for combinations, permutations, and permutations

What is the formula for calculating combinations and permutations? This article will guide you on how to calculate combinations and other related formulas.

Permutations and Combinations are the most basic concepts in mathematics that involve selecting items from a group or set.

• Permutation is the arrangement of items in order of selection from a given group.
• Combination is the selection of items without regard to order.

Table of Contents

• Combinatorial formula
• Permutation formula
• Permutation 

Combinatorial formula

Given a set A with n elements and an integer k, (1 ≤ k ≤ n). Each subset of A with k elements is called a k-fold combination of n elements of A.

K-combination formula of n

Formula for the properties of a combination:

Examples of combinatorics

Example 1: 

A group of 12 students. How many ways are there:

a) Choose 2 representatives for the group

b) Choose 2 people and assign the positions of team leader and deputy team leader.

c) Divide the group into 2 groups, in which the group leader and deputy group leader are in different groups.

Solution

a) Choose 2 friends from 12 friends who are combinations of 2 of 12: C122 = 66 ways.

b) Choose 2 people and assign them the position of combining 2 of 12: A122 = 132 ways.

c) Divide the group into 2 groups, each group has 6 members.

In which the team leader and deputy team leader are in different groups.

Choose 5 friends to be in the same group as the team leader from the remaining 10 friends: C105 = 252 ways.

Choose 5 people to be in the same group as the deputy leader from the remaining 5 people: C55 = 1 way.

So there are 252.1 = 252 ways.

Permutation formula

Given a set A with n elements and an integer k, (1 ≤ k ≤ n). When we take k elements of A and arrange them in an order, we get a k-fold permutation of n elements of A (called an n-fold permutation of k of A).

The number of k-permutations of a set with n elements is:

Permutation formula:

• Some conventions: 0! = 1, An0 = 1, Ann = n!
• Characteristics: This is an ordered sort and the number of elements to be sorted is k: 0 ≤ k ≤ n.

For example: 

From the digits 0 to 9. How many ways are there to form a natural number such that:

a) Number with 6 different digits

b) A number with 6 different digits and divisible by 10

c) Odd numbers have 6 different digits.

Solution

a) Make a number with 6 different digits

Choose the first digit from numbers 1 to 9: there are 9 ways to choose

The remaining digits are the 5th permutation of the remaining 9 numbers (other than the first digit) with A95

So there are 9A95 = 136080 numbers.

b) A number with 6 different digits and divisible by 10

Choose the unit digit: there is 1 way to choose the digit 0

Choose the remaining digits as the 5th permutation of the remaining 9 numbers (other than the digit 0) with A95

So there are A95 = 15120 numbers.

c) Let the number be an odd number with 6 different digits made from digits 0 to 9.

Because it is an odd number, f ∈{1; 3; 5; 7; 9}

Choose f: there are 5 ways to choose

Choose a from the digits {1; 2; 3; 4; 5; 6; 7; 8; 9}\{f}: there are 8 ways to choose

Choose b, c, d, e as the 4-complex of the remaining 8 digits (other than f and a): we have A84

So there are 5.8A84 = 67200 numbers.

Permutation

a) Definition:

- Given a set A of n elements (n ≥ 1).

Each result of an ordering of n elements of a set A is called a permutation of n elements.

- Note: The two permutations of n elements differ only in their arrangement order.

b) Number of permutations:

- The symbol Pn is the number of permutations of n elements.

Permutation formula:

Pn = n(n – 1)…2.1 = n!

Convention: 0! = 1; 1! = 1.

For example:  Arrange 10 people, including 5 boys and 5 girls, on a bench. How many ways are there to arrange so that:

a) Sort any

b) The boys sit next to each other

c) Boys and girls sit alternately.

Solution

a) The number of ways to arrange 10 people on a bench is a permutation of 10: 10!

b) Arrange the boys to sit next to each other. We put 5 boys into a "bundle": there are 5! ways to arrange inside the "bundle"

Then arrange 5 girls together in a "bunch" on a bench. There are: 6! ways to arrange.

So there are 5! . 6! = 86400 ways to arrange the boys to sit next to each other.

c) Suppose 10 people are arranged on benches numbered from 1 to 10.

To alternate between boys and girls

+ Case 1: Boys sit in odd positions, girls sit in even positions

Number of ways to arrange the boys: 5!

Number of ways to arrange girls: 5!

Therefore there are 5! . 5! ways to arrange.

+ Case 2: Boys sit in even positions, girls sit in odd positions

Similar to the above case, we have 5! . 5! ways to arrange.

So there are 2 . 5! . 5! = 28800 ways to arrange.

Difference between permutation and combination

The difference between permutation and combination can be understood through the following table:

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